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The single feed mode has become very important in systems powered from a single power supply, because it reduces the cost of circuits with operational amplifiers and also makes it easier to use on mobile devices. In this series of articles will look at various ways that can power amplifiers circuits using operational amplifiers with single power supply (single rail).

The diagram of a sound amplifier, which operates with a single power supply shown in Figure 16. In this circuit one end of the signal source connected to the right input terminal of the operational amplifier, and the other end connected to the bias voltage VBias.

The load resistance (RL) and R1 also associated with the bias voltage. The voltage at the right input terminal, VP, is equal to:


The voltage at the reverse input terminal VN, is equal to:

The bias voltage (VBias) is typically equal to half the voltage of the power supply. The output voltage of this amplifier can be calculated as follows:

The current I1, which flows through the resistor R1, is approximately equal to the current I2, which flows through the resistor R2, and thus I1 = I2. The current flowing through R1 is equal to the difference voltage across R1 divided by the value of R1, ie


In this circuit VB is the voltage applied to the right end of R1, and VA is the voltage applied to the left end of R1.


Similarly, the current I2 flowing through the resistor R2, is equal to the difference voltage across the R2.






By multiplying both sides of this equation with R2 we get:


So in the absence of an input signal, ie when VIN = 0, the above equation becomes


So in the absence of an input signal, the input terminals and the output terminal of the operational amplifier have the same voltage (VBias).

If an input voltage 2 volts peak to peak, is applied to the amplifier shown in Figure 16, R1 = R2, and VCC = 10V, then when the input signal is at the positive peak value of +1V, the output voltage will be:





When the input signal changes is at its highest negative value -1V, then the output voltage equals:

 So when the input signal changes from -1V to +1V, the output voltage varies from +3V to +7V.

Figure 17 shows waveforms of the input and output. Notice that the first term in equation 1 represents the voltage gain of a conventional amplifier using a double power supply. The voltage gain is given by the following relationship:


In a conventional non-inverting amplifier, which operates with double supply, the bias voltage VBias is equal to 0 volts, and the output voltage is given by the following equation:

FM Radio Receiver with IC TDA 7012T

From the picture above components to make the FM Radio Receiver IC TDA 7012T as follows:

R1 = 8kΩ2

R2 = 10kΩ

R3 = 390Ω

C1, C3 = 10nF

C2, C6, C9, C16 = 100nF

C4 = 33pF

C5 = 25pF trimmer

C7, C10 = 1nF5

C8 = 820pF C11 = 1NF

C12 = 68pF

C13 = 220pF

C14 = 47μF 10V

C15 = 3nF3

L1 = 36nH

L2 = 1μH,

IC1 = TDA7021T

For a collector follower with emitter resistor, you’ll often find that the gain per stage is no more than 10 to 50 times. The gain increases when the emitter resistor is omitted. Unfortunately, the distortion also increases. With a ubiquitous transistor such as the BC547B, the gain of the transistor is roughly equal to 40 times the collector current (Ic), provided the collector current is less than a few milliamps. This value is in theory equal to the expression q/KT, where q is the charge of the electron, K is Boltzmann’s constant and T is the temperature in Kelvin.

For simplicity, and assuming room temperature, we round this value to 40. For a single stage amplifier schema with grounded emitter it holds that the gain Uout /Uin (for AC voltage) is in theory equal to SRc. As we observed before, the slope S is about 40Ic. From this follows that the gain is approximately equal to 40I cRc. What does this mean? In the first instance this leads to a very practical rule of thumb: that gain of a grounded emitter schema amounts to 40·I c·Rc, which is equal to 40 times the voltage across the collector resistor.

If Ub is, for example, equal to 12 V and the collector is set to 5V, then we know, irrespective of the values of the resistors that the gain will be about 40R(12–5) = 280. Notable is the fact that in this way the gain can be very high in theory, by selecting a high power supply voltage. Such a voltage could be obtained from an isolating transformer from the mains. An isolating transformer can be made by connecting the secondaries of two transformers together, which results in a galvanically isolated mains voltage.

10,000x With One Transistor Circuit diagram

That means, that with a mains voltage of 240 Veff there will be about 340 V DC after rectification and filtering. If in the amplifier schema the power supply voltage is now 340 V and the collector voltage is 2 V, then the gain is in theory equal to 40 x (340–2). This is more than 13,500 times! However, there are a few drawbacks in practice. This is related to the output characteristic of the transistor. In practice, it turns out that the transistor does actually have an output resistor between collector and emitter.

This output resistance exists as a transistor parameter and is called ‘hoe’. In normal designs this parameter is of no consequence because it has no noticeable effect if the collector resistor is not large. When powering the amplifier from 340 V and setting the collector current to 1 mA, the collector resistor will have a value of 338 k. Whether the ‘hoe’-parameter has any influence depends in the type of transistor. We also note that with such high gains, the base-collector capacitance in particular will start to play a role.

As a consequence the input frequency may not be too high. For a higher bandwidth we will have to use a transistor with small Cbc, such as a BF494 or perhaps even an SHF transistor such as a BFR91A. We will have to adjust the value of the base resistor to the new hfe. The author has carried out measurements with a BC547B at a power supply voltage of 30 V. A value of 2 V was chosen for the collector voltage. Measurements confirm the rule of thumb. The gain was more than 1,000 times and the effects of ‘hoe’ and the base-collector capacitance were not noticeable because of the now much smaller collector resistor. Link

This is a Simple Comparator with time out Circuit Diagram. The MC1422 is used as a comparator with input (Pin 5). The frequency of the pulses for the capability of a timing output pulse when the the values of R2 and Cl as shown is approx i-inverting input (Pin 6) is the non inverting mately 2 Hz, and the pulse width 0 ms.

This schema sent out an alarm when its LDR sensor is exposed to light by sun or lamp. A 555 astable multivibrator was used here which sent signal a tone of about 1kHz upon detecting light.The sensor when exposed by light completes the schema and makes the 555 oscillate at about 1kHz with transistor to drive current.

The sensor is also shown in the schema diagram. It has to placed making an angle of about 30 – 45 degrees to the ground.

Sensitivity can be adjust with P1.  This makes the sun light to flow through it to the ground and prevents the alarm from going on due to the stored light on the sensor.